Question

a) a 90% ci for test scores is (80, 87). this means that the average score for the population is most likely 83.5
b) a 90% ci for test scores is (80, 87). this means that 90% of all scores fall in this range.
c) a 90% ci for test scores is (80, 87). this means that we are 90% confident that the population mean is in this range.
d) if we double our sample size, our confidence interval width will be cut in half.
e) if we increase our confidence level, we will increase our margin of error.

Explanation:

• a) The confidence interval is about the population mean, not a single most - likely value. The mid - point of the interval \((\frac{80 + 87}{2}=83.5)\) is the sample mean (if it's a symmetric interval for a normal distribution assumption), but the interpretation of the confidence interval is not about a single most - likely value for the population mean.
• b) A confidence interval for the mean is not about the proportion of data values. It is about the range within which the population mean is likely to lie.
• c) By the definition of a confidence interval, a \(90\%\) confidence interval means that if we were to construct many such intervals from different samples, about \(90\%\) of them would contain the True population mean. So, we are \(90\%\) confident that the population mean is in the interval \((80,87)\).
• d) The width of a confidence interval for the mean (assuming a normal distribution and known or estimated standard deviation) is \(w = 2z\frac{\sigma}{\sqrt{n}}\) (for a z - interval) or \(w=2t\frac{s}{\sqrt{n}}\) (for a t - interval). If we double \(n\) (i.e., \(n_{new}=2n\)), the new width \(w_{new}=2z\frac{\sigma}{\sqrt{2n}}=\frac{w}{\sqrt{2}}

eq\frac{w}{2}\)

• e) The margin of error \(E = z\frac{\sigma}{\sqrt{n}}\) (for a z - interval) or \(E = t\frac{s}{\sqrt{n}}\) (for a t - interval). As the confidence level increases, the value of \(z\) (for a z - interval) or \(t\) (for a t - interval) increases. For example, for a 90% confidence interval, \(z = 1.645\) (approximate for a large sample z - interval), and for a 95% confidence interval, \(z=1.96\). Since \(E\) is directly proportional to \(z\) (or \(t\)), increasing the confidence level increases the margin of error.

Answer:

C. A 90% CI for test scores is (80, 87). This means that we are 90% confident that the population mean is in this range.
E. If we increase our confidence level, we will increase our margin of error.